Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mx = 105;
int g[mx][mx], n, m;
vector<pair<string, int> > eval(bool flag) {
  vector<pair<string, int> > ret;
  while (1) {
    int zero = 0;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        if (!g[i][j]) zero++;
    if (zero == n * m) break;
    bool up = false;
    for (int i = 0; i < n; i++) {
      int cur = 1e9;
      for (int j = 0; j < m; j++) {
        cur = min(cur, g[i][j]);
      }
      if (cur) {
        up = true;
        string label = "row";
        if (flag) label = "col";
        for (int j = 0; j < cur; j++) {
          ret.push_back(make_pair(label, i + 1));
        }
        for (int j = 0; j < m; j++) g[i][j] -= cur;
      }
    }
    for (int j = 0; j < m; j++) {
      int cur = 1e9;
      for (int i = 0; i < n; i++) {
        cur = min(cur, g[i][j]);
      }
      if (cur) {
        up = true;
        string label = "col";
        if (flag) label = "row";
        for (int i = 0; i < cur; i++) ret.push_back(make_pair(label, j + 1));
        for (int i = 0; i < n; i++) g[i][j] -= cur;
      }
    }
    if (!up) {
      ret.clear();
      return ret;
    }
  }
  return ret;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> g[i][j];
  int gg[mx][mx];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) gg[i][j] = g[i][j];
  vector<pair<string, int> > ret;
  for (int i = 0; i < 2; i++) {
    vector<pair<string, int> > cur = eval(i);
    int zero = 0;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        if (!g[i][j]) zero++;
    if (zero == n * m) {
      if (ret.empty())
        ret = cur;
      else {
        if (cur.size() < ret.size()) ret = cur;
      }
    } else {
      if (cur.empty()) {
        cout << -1 << endl;
        return 0;
      }
    }
    int h[m][n];
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) h[j][i] = gg[i][j];
    for (int i = 0; i < m; i++)
      for (int j = 0; j < n; j++) g[i][j] = h[i][j];
    swap(n, m);
  }
  int sz = ret.size();
  cout << sz << endl;
  for (int i = 0; i < sz; i++)
    cout << ret[i].first << " " << ret[i].second << endl;
  return 0;
}